# Full text of "Time Optimization, Fermat's Principle and Mathematics-Based Technology Enhanced Learning using Excel"

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```Time Optimization, Fermat's Principle and
Mathematics-Based Technology Enhanced
Learning using Excel

By Patrick Bruskiewich

Abstract

We can view reflection of light off a mirror, and refraction of light leading to Snell's Law
as a Time Optimization Problem (TOP). Fermat's Principle is that the path travelled by
light is the path of least travel time, an explicit description of time optimization. A third
TOP problem, flight time between two points, with auxiliary travel constraints, will also
be considered in this paper. Solving Time Optimization Problem using a spreadsheet
program like EXCEL is a good example of Mathematics-based Technology Enhanced
Learning (MTEL) using a widely available software program.

Introduction

Solving Time Optimization Problem using a spreadsheet program like EXCEL is a good
example of Mathematics-based Technology Enhanced Learning (MTEL) using a widely
available software program. It is more meaningful than lower brain and animation type
programs like for instance those used in the Carl Wieman Science Education Initiative at
the University of British Columbia.

MTEL is a more meaningful approach merely because animation is a back lobe function,
while mathematics and higher reasoning is a frontal lobe one.

1.0 - The Reflection of Light off a Mirror

Consider the path that light takes when it travels from a source to a mirror and then to
your eye. This path, ABC, can be traced out using Ray Tracing (refer to Fig. 1: The
Reflection of Light off a Mirror). The dimensions of the problem are arbitrary.

For a simple system consider a light source A which is 3 units above the mirror at
distance 0, the point at which the light ray bounces off the mirror is a B at distance x, and
your eye is at C, 3 units above the distance 3 arbitrary units along the horizontal. By
Pythagoras' Theorem the two distances AB and BC become

r i= Vx 2 +3 2

^ 2 =^/(3-x) 2 +3 2

3 units

Fig. 1: The Reflection of Light off a Mirror

The total time it takes for the light to travel from A to B and then to C is

t = t x +t 2 = J " + ^-

c c

For the sake of simplicity let c = 1, then the problem simplifies to

r = r x +r 2

EXCEL program data for this problem is given in Table 1 : Reflection Travel Time as a
Function of Horizontal Distance x. By inspection, the minimum time is at x = 1.5, with t
= 6.708204 for our system configuration. The data for the reflection travel time is
graphed in Fig.2: Total time as a function of Horizontal Distance x for Reflection. The
Extremum in travel time is at x= 1.5.

The Extremum for Reflection can be determined explicitly using Calculus,

dr dr. dr~

— = ^- + ^- =

dx dx dx

This means

2x 2(3-x)

2 V* 2 +3 2 2^/(3 -xf +3

=

Solving for horizontal distance x we find

2/o \ 2 . o2 2 2/o \2 , 2

X

(3-x) +3 2 x 2 =x 2 (3-x) +3 2 (3-*y

x 2 =(3-xy

2jc = 3
x = \.5

which is half the horizontal distance, as expected.

It is easy to show the general case with the light source and the eye at different heights
above the mirror.

X

R1

T1

R2

T2

T

3

3

4.242641

4.242641

7.242641

0.1

3.001666

3.001666

4.172529

4.172529

7.174195

0.2

3.006659

3.006659

4.103657

4.103657

7.110316

0.3

3.014963

3.014963

4.036087

4.036087

7.05105

0.4

3.026549

3.026549

3.969887

3.969887

6.996436

0.5

3.041381

3.041381

3.905125

3.905125

6.946506

0.6

3.059412

3.059412

3.841875

3.841875

6.901286

0.7

3.080584

3.080584

3.780212

3.780212

6.860796

0.8

3.104835

3.104835

3.720215

3.720215

6.82505

0.9

3.132092

3.132092

3.661967

3.661967

6.794059

1

3.162278

3.162278

3.605551

3.605551

6.767829

1.1

3.195309

3.195309

3.551056

3.551056

6.746365

1.2

3.231099

3.231099

3.498571

3.498571

6.72967

1.3

3.269557

3.269557

3.448188

3.448188

6.717744

1.4

3.310589

3.310589

3.4

3.4

6.710589

1.5

3.354102

3.354102

3.354102

3.354102

6.708204

1.6

3.4

3.4

3.310589

3.310589

6.710589

1.7

3.448188

3.448188

3.269557

3.269557

6.717744

1.8

3.498571

3.498571

3.231099

3.231099

6.72967

1.9

3.551056

3.551056

3.195309

3.195309

6.746365

2

3.605551

3.605551

3.162278

3.162278

6.767829

2.1

3.661967

3.661967

3.132092

3.132092

6.794059

2.2

3.720215

3.720215

3.104835

3.104835

6.82505

2.3

3.780212

3.780212

3.080584

3.080584

6.860796

2.4

3.841875

3.841875

3.059412

3.059412

6.901286

2.5

3.905125

3.905125

3.041381

3.041381

6.946506

2.6

3.969887

3.969887

3.026549

3.026549

6.996436

2.7

4.036087

4.036087

3.014963

3.014963

7.05105

2.8

4.103657

4.103657

3.006659

3.006659

7.110316

2.9

4.172529

4.172529

3.001666

3.001666

7.174195

3

4.242641

4.242641

3

3

7.242641

Table 1: Reflection Travel Time as a function of horizontal distance x

7.3
7.2
7.1
7
6.9
6.8
6.7
6.6
6.5
6.4

13 5 7

Reflection Law

11 13 15 17 19 21 23 25 27 29
0< x<3

Fig.2: Total time as a function of Horizontal Distance x for Reflection

The angle to the vertical at the reflection point is

X

(3-x)

= sin X

- sin #,

By inspection we find for the two angles to the vertical at the reflection point that

9 X =6 2

which is easily validated experimentally.

2.0 Refraction through a Medium with an Index of Refraction n

Now consider light traveling from one medium with speed vi, through an interface, into
another medium with a different speed V2 (refer to Fig. 2: The Refraction of Light in
Two Media). The dimensions of the problem are arbitrary.

For a simple system consider a light source D which is 3 units above the interface at
distance 0, the point at which the light ray enters the second medium is a E at distance x,
and your eye is at F, 3 units below the distance 3 arbitrary units along the horizontal.

3 units

medium 1: vl

3 units

medium 2: v2

Fig. 3: The Refraction of Light in Two Media

By Pythagoras' Theorem the two distances DE and EF become

r x = yjx 2 + 3 2

^=V( 3_x ) 2+32

A representative example is that of light traveling from air into water (light travels at a
speed 0.75 in water compared to air) EXCEL program data for this problem is given in
Table 2: Travel Time as a Function of Horizontal Distance x for Refraction. The data
for the reflection travel time is graphed in Fig.4: Total time as a function of Horizontal
Distance x for Refraction.

The total travel time for the refraction problem is

t = t 1 +t 2 =— +

v x v 2

The Extremum for our refraction problem where V2 = 0.75 vi is at some point between
1.7 and 1.8 for our system configuration.

To find the Extremum using Calculus we see that

dr
dx

f 1 > \

\ v u

dr,

dx

r 1 s \

\ V 2J

dn

dx

=

From whence we have

'P

2x

vj2ylx 2 +3

'O

\ V U

2(3-x)

2 . o2

2J(3-x) +3

R1

T1

R2

T2

T

0.1

3.001666

3.001666

4.172529

5.563372

8.565038

0.2

3.006659

3.006659

4.103657

5.471543

8.478202

0.3

3.014963

3.014963

4.036087

5.38145

8.396412

0.4

3.026549

3.026549

3.969887

5.293182

8.319731

0.5

3.041381

3.041381

3.905125

5.206833

8.248214

0.6

3.059412

3.059412

3.841875

5.122499

8.181911

0.7

3.080584

3.080584

3.780212

5.040282

8.120867

0.8

3.104835

3.104835

3.720215

4.960287

8.065122

0.9

3.132092

3.132092

3.661967

4.882622

8.014714

1

3.162278

3.162278

3.605551

4.807402

7.969679

1.1

3.195309

3.195309

3.551056

4.734742

7.930051

1.2

3.231099

3.231099

3.498571

4.664762

7.89586

1.3

3.269557

3.269557

3.448188

4.597584

7.86714

1.4

3.310589

3.310589

3.4

4.533333

7.843922

1.5

3.354102

3.354102

3.354102

4.472136

7.826238

1.6

3.4

3.4

3.310589

4.414119

7.814119

1.7

3.448188

3.448188

3.269557

4.359409

7.807597

1.8

3.498571

3.498571

3.231099

4.308132

7.806703

1.9

3.551056

3.551056

3.195309

4.260412

7.811468

2

3.605551

3.605551

3.162278

4.21637

7.821921

2.1

3.661967

3.661967

3.132092

4.176123

7.838089

2.2

3.720215

3.720215

3.104835

4.13978

7.859995

2.3

3.780212

3.780212

3.080584

4.107446

7.887657

2.4

3.841875

3.841875

3.059412

4.079216

7.92109

2.5

3.905125

3.905125

3.041381

4.055175

7.9603

2.6

3.969887

3.969887

3.026549

4.035399

8.005286

2.7

4.036087

4.036087

3.014963

4.01995

8.056037

2.8

4.103657

4.103657

3.006659

4.008879

8.112536

2.9

4.172529

4.172529

3.001666

4.002222

8.174751

3

4.242641

4.242641

3

4

8.242641

Table 3: Travel Time as a Function of Horizontal Distance x for Refraction

In terms of ij and V2 is merely

(\ \

v v iy

x J l)(3-x)

\ V U

The orientation of the rays traveling through the two media is best done with angles taken
with respect to the perpendicular to the interface.

Snell's Law

12 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 2021222324252627282930

< x<3

Fig.4: Total time as a function of Horizontal Distance x for Refraction

By inspection then we see that

— = sin 6 X

(3-jc)

= sin 0^

If we express the Extremum condition in terms of angles and also multiply both sides by
the speed of light c we arrive at Snell's Law, namely

n x sin l = n 2 sin 2

where the index of refraction for the two media are

'<0

n x =

V V J

( c )

n 2 =

UJ

For instance, in our example for the second medium, v 2 = 0.75 vi and n 2 = 1.33. It is
evident then that Snell's Law, a Time Optimization problem, is a direct result of Fermat's
Principle.

3.0 Flight Time Between Points with Auxiliary Travel Constraints

Consider flight between two points that involves an auxiliary travel constraint. The flight
constraint may be fuel expenditure, or perhaps a different risk of bad weather or variable
energy expenditure due to flight conditions across water compared to land. The
constraint might be a higher risk of being downed by a Surface to Air Missile (SAM) if
the problem were a combat simulation over hostile territory involving a "Hot Zone".

If the flight expenditure over water is s l while that over land is s 2 , with s l < s 2 , then an
optimal flight path needs to be determined by means of the expenditure ratio R defined as

R = ^

The flight path is given in Fig. 5: Flight Between Two Points with an Auxiliary Travel
Constraint. The direct route GK is not the preferred route. The modified route GJK
becomes the preferred route.

H

he

K

Fig. 5: Flight Between Two Points with an Auxiliary Travel Constraint

The distance GH is given by r. The distance GJ is

GJ =

The distance HJ is

The distance JK is then

sin#

HJ = r coiO

JK = s - r cot

The full expenditure E of the flight path is given by

E = s x n3J + s 2 UJK

y

= s x 0— - 1- s 2 \\s-r cot 0)

sin#

= £ 2 Us + £ r

- r cot
J

sin#

The bracketed term is the only term that depends on the angle, therefore it is this term
that we minimize. Define w such that

( R

w

V

sin#

-rcotO

\

J

Using Calculus, we find the Extremum in w

dw l-R cos 6

d6 sin 2 6

From which we solve for the angle (recall s l = Rs 2 with 1 < R )

R cos - 1 =^> - cos

EXCEL program data for this flight problem is given in Fig 6: The Optimal Flight Angle
as a function of R. From the stand point of initial conditions there is a minimal angle <fi

(j) = tan"

GH
HK

For expenditure ratio R=1.2 the optimum angle is 33 degrees while for R=2 the angle is
60 degrees.

If 1 □ R then the best strategy is to fly GH then HK that is » 90° .

Fig 6: The Optimal Flight Angle as a function of R

4.0 Using EXCEL - Technology Enhanced Learning

Solving Time Optimization Problem such as the three examples outlined in this paper
using a spreadsheet program like EXCEL, is a good example of Mathematics -based
Technology Enhanced Learning (MTEL).

based computers. After outlining the general concept to the problem, the instructor can
invite the students to complete the assignment with a minimum of assistance.

Pre-packaged software programs like Matlab puts a distance between the student and the
mathematical algorithms in use by the program. Using EXCEL would require the student
to explicitly write the algorithm themselves before attempting the problem.

Using a spreadsheet program like EXCEL as a Mathematics-based Technology Enhanced
Learning (MTEL) problem is a more meaningful approach to learning than lower brain
teaching practices like the animation type programs and relating cartoon like materials
used in the Carl Wieman Science Education Initiative at the University of British
Columbia. Such animation is a back lobe function

MTEL is a more meaningful approach because mathematics and higher reasoning is a
frontal lobe one. The author has mentioned to Carl Wieman a number of times that the
greatest distance that most physics students in his program have to travel is between the
occipital lobe, where vision information and animation is interpreted, to their frontal lobe.
Why not start in the frontal lobe with mathematics and higher reasoning? A student's
brain is a noisy place with many distractions. The occipital lobe is under developed
while their frontal lobes are not.

The Carl Wieman Science Education Initiative at the University of British Columbia has
fallen far short of pedagogical expectations partly because it was set in place by
and a very limited understanding of quality professional pedagogical practice. The Carl
Wieman Science Education Initiative at the University of British Columbia is somewhat
simplistic in it is constructs and should be wound down, and the savings in public funding
spent elsewhere to improve physics education in the Province of BC.

Sadly over the past five years many students who have gone through the Carl Wieman
Science Education Initiative at the University of British Columbia have missed an
opportunity to more fully develop their skills in both basic and applied physics. It is yet
to be determined what such a shortfall will ultimately cost the Province and People of
British Columbia. Perhaps this shortfall should be taken out of the salaries of the
administrative mandarins ... at penny ante on a dollar lost?